Dot Product of Constant Magnitude Vector-Valued Function with its Derivative is Zero

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Theorem

Let:

$\map {\mathbf f} x = \ds \sum_{k \mathop = 1}^n \map {f_k} x \mathbf e_k$

be a differentiable vector-valued function.


Let $\map {\mathbf f} x$ be such that its magnitude is constant:

$\size {\map {\mathbf f} x} = c$

for some $c \in \R$.


Then the dot product of $\mathbf f$ with its derivative is zero:

$\map {\mathbf f} x \cdot \dfrac {\d \map {\mathbf f} x} {\d x} = 0$


Proof 1

\(\ds \map {\mathbf f} x \cdot \dfrac {\d \map {\mathbf f} x} {\d x}\) \(=\) \(\ds \size {\map {\mathbf f} x} \dfrac {\d \left\lvert{\map {\mathbf f} x}\right\rvert} {\d x}\) Dot Product of Vector-Valued Function with its Derivative
\(\ds \) \(=\) \(\ds \size {\map {\mathbf f} x} \dfrac {\d c} {\d x}\) Magnitude of $\mathbf f$ is constant
\(\ds \) \(=\) \(\ds \size {\map {\mathbf f} x} \times 0\) Derivative of Constant


Hence the result.

$\blacksquare$


Proof 2

\(\ds \map {\mathbf f} x \cdot \map {\mathbf f} x\) \(=\) \(\ds c^2\) Given assumption
\(\ds \dfrac {\d f} {\d x} \paren {\map {\mathbf f} x \cdot \map {\mathbf f} x}\) \(=\) \(\ds 0\) Derivative of Constant
\(\ds \dfrac {\d \map {\mathbf f} x} {\d x} \cdot \map {\mathbf f} x + \map {\mathbf f} x \cdot \dfrac {\d \map {\mathbf f} x} {\d x}\) \(=\) \(\ds 0\) Derivative of Dot Product of Vector-Valued Functions
\(\ds 2 \map {\mathbf f} x \cdot \dfrac {\d \map {\mathbf f} x} {\d x}\) \(=\) \(\ds 0\) Dot Product Operator is Commutative
\(\ds \map {\mathbf f} x \cdot \dfrac {\d \map {\mathbf f} x} {\d x}\) \(=\) \(\ds 0\)

$\blacksquare$


Sources

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